Recent content by Swatje

Ok. Thanks for your reply. Let's consider two situations: 1) You have 10 numbers, and one hops out. What you do is you calculate s, you calculate the average, then you look at difference between the average and the "one hopping out", divide by s and then look at the t-value... t =...

Hi, I was wondering, if you have a certain sample set. How can you quantify the probability of a certain statistical fluctuation? For example, let's say you have a poisson distribution of high numbers, up into the fifties. You have one peak... How do you determine the probability of this being...

Nah, a coursework question. But I understand it now, case closed ;).

Hello there!I have a small question regarding Heisenberg uncertainty relationship. If one sets up a double slit experiment and one looks through which hole the electron passes, so it acts as a particle and you know exactly where it is. If immediately after that slit, you place some sort of...

Im afraid your going to need a lot of tesla's to construct it. 1 maybe 1.5 tesla.

Yep thx, and I also forget to add a 10 of the amount of coils.

Oh yes i see, small mistake ;). so then: I = - (m g delta h ) / (L * 2r * B) and delta h, will approxiamtely be: - sin theta * PI * r * 1/2 so I = (m*g*pi) / ( 4 * L * B) And you can discard any increase by faraday, because it will eventually have to stop rolling anyways, and then I = I...

Yes, at time = 0 the netforces cancel each other out, but when it starts rolling, the forces won't be strictly radial anymore and there will be a tangential component which exerts torque. I understand that magnetic force does no work, but can't it oscillate slightly over an angle smaller than...

So Now the minimal amount of I, would be for omega*t to be as large as possible right? So let's say P = period/4, so after 90 degrees: mgh = - M . B mg*sin(theta)*s = - I * 2*r*L * B * cos (omega*t) I = - (m*g*sin(theta)*s)/(2*r*L*B*cos(omega*t)) Now we want I as small as possible...

Ok, i had a small idea here, tell me if its mathematically correct: If the cilinder starts rolling, let's say with at angle theta, let's let it roll over a certain height difference... 1/2*mv² + 1/2*I*omega = mgh mv² + mr²omega² = 2mgh v² + r²omega² = 2gh 2*r²omega² = 2gh (Last...

Okay, quick addition: Electric field of a uniformely charged, infinitely long rod: E = L'/(2*pi*epsilon-0*R') where L' is the amount of charges per meter of the rod. If you would transform this to a fast observer, or the current moving fast, you would get several distortions: a) the amount...

Well I'll help you let me type. Okay. First of all its a law for currents and currents only. These must not be finited currents otherwise they would be observed as a row of sinlge moving charges. You'll see later on that you can't use biot-savart on a single charge. I'm going to explain...

Depends, either let it decline radially through Newton's law, or you could use a rotating rocket around the earth, and try and calculate its minimum height with corresponding speed to rotate around earth, using gravitational acceleration as a centripetal force.

I didnt check. But you didnt have any concrete data to start on, but it seemed plausible.

Yes you could do that and calculate it. But that's not the purpose of the exercise. The purpose is to look at G. G = 6.67 * 10^-11. This means 0.0000... A very small number. And just by looking at this small number you should be able to guess that it's probably not the case, just as we see in...